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Let $E$ be a smooth vector bundle equipped with an affine connection $\nabla$.

Suppose that $(E,\nabla)$ admits a non-zero parallel section. I think that $(\bigwedge^k E,\bigwedge^k \nabla)$ does not need to admit a non-zero parallel section even locally. How to construct such an example?

This seems especially interesting when $k < \text{rank}(E)$.

Moreover generally , are there any non-trivial relations between the dimension of the space of parallel sections of $(E,\nabla)$, and that of $(\bigwedge^k E, \bigwedge^k\nabla)$ (locally)?

If there are $k$ independent parallel sections of $E$, then $\bigwedge^k E$ has at least one parallel section; $\sigma_1,\dots,\sigma_k$ are parallel $\Rightarrow \sigma_1 \wedge \dots \wedge \sigma_k$ is parallel.

What happens if $E$ has $r<k$ parallel sections? Does $\bigwedge^k E$ still admit (locally) a non-zero parallel section?

Edit: We can probably use the relation between the curvatures: Let $X,Y \in \Gamma(TM)$. Then $R^{ \bigwedge^k\nabla}(X,Y)=d\psi_{\operatorname{Id}} (R^{\nabla}(X,Y)) $, where $\psi:\text{End}(E) \to \text{End}(\bigwedge^k E)$ is the exterior power map, $\psi(A)=\bigwedge^k A$.

Since the dimension of the space of local parallel sections around $p \in M$ equals $\ker R(\cdot,\cdot)$, it suffices to construct an example where $R^{\nabla}(\cdot,\cdot)$ is singular, but $R^{ \bigwedge^k\nabla}(\cdot,\cdot)$ is invertible. This is certainly possible on the algebraic level, see e.g. example 1, in this question, with $k=2, \text{rank}E=3$.

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  • $\begingroup$ So, certainly when $k=\text{rank}(E)$ you can't have a nowhere-vanishing section of $\Lambda^k E$ unless $E$ is an orientable bundle. I suspect you can create examples in general by taking $E=F\oplus L$ with $L$ a trivial line bundle (with an obvious split connection). $\endgroup$ – Ted Shifrin Feb 10 at 22:19
  • $\begingroup$ Thank you. I am actually more interested in the local problem, but I forgot to mention this in the question. My apologies. (I have now edited the question accordingly). $\endgroup$ – Asaf Shachar Feb 11 at 7:57
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The answer to your question is no.

Fix a manifold $M$ of dimension $m \geq 2$ and let $E$ be a trivial rank two bundle over $M$. Since your question is local, we might as well assume that $M$ is just an open subset of $\mathbb{R}^m$ but this won't simplify anything. Let $(e_1,e_2)$ be a global frame for $E$ over $M$ and choose some non-closed one form $\alpha \in \Omega^1(M)$. Define a connection $\nabla = \nabla^{\alpha}$ on $E$ by the formula

$$ \nabla_X(f^1 e_1 + f^2 e_2) = (Xf^1)e_1 + ((Xf^2) + \alpha(X)f^2)e_2. \tag{1}$$

You can readily check that this indeed defines a connection and it satisfies

$$ \nabla_X(e_1) \equiv 0, \nabla_X(e_2) = \alpha(X) e_2,\\ R_{\nabla}(X,Y)(f^1 e_1 + f^2 e_2) = f^2 \cdot d\alpha(X,Y) \cdot e_2. $$

In particular, $R_{\nabla} \neq 0$ so $E$ doesn't have two independent parallel sections with respect to $\nabla$, only one (up to a constant multiple): $e_1$. Let's check whether $\Lambda^2(E)$ has a non-trivial parallel section. A general section of $\Lambda^2(E)$ has the form $g(e_1 \wedge e_2)$ where $g$ is a smooth function. Then

$$ \nabla_X(g(e_1 \wedge e_2)) = (Xg)(e_1 \wedge e_2) + g(\nabla_X(e_1 \wedge e_2)) = (Xg)(e_1 \wedge e_2) + g(\nabla_X e_1 \wedge e_2 + e_1 \wedge \nabla_X e_2) = (Xg + g\cdot \alpha(X))e_1 \wedge e_2.$$

Hence, $\Lambda^2(E)$ has a non-trivial parallel section if and only if we can find a non-zero $g$ which satisfies the equation

$$ dg + g \alpha = 0. \tag{2}$$

This is a partial differential equation for $g$. Wedging the equation with $\alpha$ and taking into account that $\alpha$ is a one-form, we get $$ dg \wedge \alpha + g \alpha \wedge \alpha = dg \wedge \alpha = 0.$$ Taking the exterior derivative of the equation, we get

$$ 0 = d^2g + dg \wedge \alpha + g d\alpha = g d\alpha. $$

Since we are looking for non-zero $g$, we see that a necessary condition for the equation to have a solution is $d\alpha = 0$ which doesn't hold since we assumed $\alpha$ is not closed.

In fact, for the family of connections $\nabla^{\alpha}$ we see that $\nabla^{\alpha}$ has two linearly independent parallel sections if and only if the induced connection on $\Lambda^2(E)$ has one linearly independent parallel section. The curvature of the induced connection on $\Lambda^2(E)$ can be identified simply with $d\alpha$. Indeed, $$R^{ \bigwedge^2\nabla}(X,Y)(e_1 \wedge e_2)=R^{ \nabla}(X,Y)(e_1) \wedge e_2+e_1 \wedge R^{ \nabla}(X,Y)(e_2)=d\alpha(X,Y)e_1 \wedge e_2.$$

Since $\Lambda^2(E)$ is a line bundle, it admits a non-zero parallel section if and only if its curvature tensor $R^{ \bigwedge^2\nabla}$ vanished identically, i.e. if and only if $\alpha$ is closed.

Also, note that formula $(1)$ implies that $Xg + g\cdot \alpha(X)=0$ if and only if $\nabla_X(g e_2)=0 $. This gives an alternative proof for the fact that equation $(2)$ admits a non-zero solution if and only if $\nabla$ is flat.

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No, this is far from true, even in the special case of connections $\nabla$ on the tangent bundle $E = TM$.

To construct an explicit example, take flowbox coordinates adapted to the $\nabla$-parallel vector field $X$, i.e., so that $X = \partial_x$, which imposes exactly that the Christoffel symbols satisfy $\Gamma_{xj}^\ell = 0$ for all $j, \ell$. In the case $\dim M = 2$, substituting these identities in the formula from this previous answer of mine to a related question of yours shows that the curvature of the connection $\nabla^2$ induced on $\Lambda^2 TM$ is $$R^2 = (-\partial_y \Gamma_{xy}^y + \partial_x \Gamma_{yy}^y) dx \wedge dy ,$$ but that answer also shows that $\Lambda^2 TM$ admits a parallel section iff $R^2 = 0$, and we can choose $\Gamma_{xy}^y, \Gamma_{yy}^y$ for which $R^2 \neq 0$. It follows quickly from that this an analogous conclusion applies to $\bigwedge^n T(M \times \Bbb R^{n - 2})$, giving counterexamples for arbitrary dimension $k = \operatorname{rank} E \geq 2$. With some CAS help it's easy to produce counterexamples for $E = TM$ and $1 < k < \dim M$, too. (Of course, these cases are more computationally involved, as the curvatures there are sections of $\Lambda^2 T^*M \otimes \operatorname{End} \Lambda^k E$, and for $0 < k < \operatorname{rank} E$ we have $\operatorname{rank} \operatorname{End} \Lambda^k E > 1$.)

There are special cases, however, in which the claim is true. For example (working locally) on a Riemannian manifold $(M, g)$, an orientation determines a Hodge star operator $\ast$ on multivectors, and since any metric connection $\nabla$ preserves $g$ and the orientation, it preserves $\ast$, too. Thus, if $X$ is a nonzero parallel section of $TM$, $\ast X$ is a nonzero section of $\bigwedge^{k - 1} T^*M$ parallel with respect to the connection induced by $\nabla$ on that bundle.

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